Cambridge A Level Mathematics - Further 9231 — 2023 May/June Paper 2 · Variant 2

1 (a) Show that the system of equations x + 2y + 3z = 1, 4x + 5y + 6z = 1, 7x + 8y + 9z = 1, does not have a unique solution. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Show that the system of equations in part (a) is consistent. Interpret this situation geometrically. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a)       5 1 2 3 4 5 6 5 2 4 3 4 7 7 9 8 6 8 9 9 6 8 7            3 12 9 0    obtain e.g. 2 3 1, 2 1, x y z y z      2 1(b) 2 3 1, 4 5 6 1, 7 8 9 1, x y z x y z x y z            1 3 1 2 2 1 z x y y x        M1 A1 Uses all three equations to reduce to one equation with two unknowns. Reducing to two equations scores M1 A0. The three planes form a sheaf. B1 Accept clear sketch or the three planes intersect along a common line. 3

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2 Use the substitution z = x + y to find the solution of the differential equation d y 1 + 3x + 3y = d x 3x + 3y - 1 for which y = 0 when x = 1. Give your answer in the form a ln ( x + y) + b ( x - y) + c = 0 , where a, b and c are constants to be determined. [7] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 d dz 1 d d y x x   B1 d 1 3 1 d 3 1 z z x z    M1 Substitutes and derives first order separable equation. d 1 3 6 1 d 3 1 3 1 z z z x z z      A1 1 1 1 2 6 d 1d z z x      M1 Separates variables and integrates both sides. 1 1 1 1 1 2 6 2 2 6 ln ln( ) z z x C x y x y C         A1 1 2 C  M1 Substitutes initial conditions into their expression.   1 1 1 6 2 2 ln( ) 0 x y x y      A1 OE. 7

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3 (a) By considering the binomial expansion of ( z + z -1 ) 4 , where z = cos i + i sin i , use de Moivre’s i + 4 cos 2i + 3) . [5] theorem to show that cos 4 i = 18 ( cos 4 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 (b) Use the substitution x = sin i to find the exact value of 2 ( 1 - x 2 ) 23 d x . [3] y0 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 1 2cos z z     B1 Use of 1 2cos z z     .       4 1 4 4 2 2 4 6 z z z z z z          M1 A1 Expands and groups. M1 A0 for no clear grouping. Correct substitution of n co is s i nz n n     for each term scores M1 A1.     4 2cos 2cos4 4 2cos2 6       M1 Substitutes 2cos . n n z z n    If n co is s i nz n n     used then must cancel sin.   4 1 8 cos cos4 4cos2 3       A1 AG. SC B1 expands   4 i c s os in    and uses trigonometric identities. 5 3(b)   1 1 3 1 2 6 2 6 π π 1 8 0 2 1 4 4 0 0 1 d cos d sin 4 2sin 2 3 x x                M1 A1 Applies substitution (M1) gets to 4 cos d   , changes limits, integration correct (A1).   9 1 16 4 3 π  A1 3

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= ( 1 + x ) d x .4 The integral In is defined by I n 5 n 0 d 5 n (a) By considering x ( 1 + x ) or otherwise, show that d x ` j, ( 5n + 1) I n n = 2 + 5 nI n -1 . 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(b) Find the exact value of I3. 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Mark scheme: 4(a)       1 5 5 5 5 d 1 5 1 1 d n n n x x nx x x x       M1 A1 Uses the product rule to differentiate.      1 5 5 5 5 1 1 1 1 n n n x x x       M1* Uses 5 5 1 1 x x  .   0 5 1 1 1 5 5 n n n n x x nI nI I            DM1 Integrates both sides using the limits given. Requires previous method mark. 1 1 2 (5 1) 5 (5 1) 2 5 n n n n n n n I nI n I nI          A1 Substitutes limits and rearranges. AG. Alternative method for question 4(a)       1 1 1 5 0 0 1 5 0 5 5 1 d 1 5 1 d n n n nI x x x x n x x x                M1 A1 Integrates by parts.       1 1 5 5 5 1 1 0 0 0 1 5 1 d 5 1 d n n n nI x x n x x n x x                M1* Uses 5 5 1 1 x x  .   1 1 0 5 1 5 5 n n n n I x x nI nI            DM1 Forms recursive formula. Requires previous method mark. 1 (5 1) 2 5 n n n n I nI     A1 Substitutes limits and rearranges. AG. 5 Question Answer Marks Guidance 4(b) 6 1 7 1 1 6 6 0 I x x        or 0 1 I  B1 2 47 2 1 2 33 11 2 10 I I I     M1 A1 Applies reduction formula. 6 2 3 323 3 3 17 16 2 15 I I I     A1 4

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5 The matrix A is given by 18 5 - 11 A = 8 6 - 4 f32 10 - 20p. (a) Show that the characteristic equation of A is m 3 - 4m 2 - 20m + 48 = 0 and hence find the eigenvalues of A. 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(b) Find a matrix P and a diagonal matrix D such that A 5 = PDP -1 . 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Mark scheme: 5(a) 18 5 11 8 6 4 0 32 10 20          6 4 8 4 8 6 18 5 11 10 20 32 20 32 10                 A1 Expands determinant. Can use other rows/columns. 2 3 4 20 0 48       A1 AG ( 2)( 4)( 6) 0        leading to 2, 4, 6    B1 4 Question Answer Marks Guidance 5(b) 24 1 2: 16 5 11 24 1 8 4 4 24 1                           i j k  M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. Must attempt to solve equations for M1. 120 1 4: 8 10 4 0 ~ 0 32 10 16 240 2                      i j k 20 1 6: 12 5 11 40 ~ 2 8 0 4 40 2                       i j k A1 A1 Thus 1 1 1 1 0 2 1 2 2            P and 32 0 0 0 1024 0 0 0 7776             D M1 A1 Or correctly matched permutations of columns. M0 if a column of zeros appears in P. 6

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6 Find the particular solution of the differential equation d 2 x d x - 12 + 36 x = 37 sin t , 2 d t d t dx given that, when t = 0, x = = 0 . [11] dt .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... 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Mark scheme: 6 2 12 36 0 m m    M1 Auxiliary equation.   6e t x At B   A1 Complementary function. Allow “ x ” missing. sin cos ' cos sin '' sin cos x p t q t x p t q t x p t q t         M1 A1 Particular integral and its derivatives. 12 36 37 p q p     12 36 0 q p q    M1 Substitutes and equates coefficients. Must have two unknowns. 35 37 p  12 37 q  A1   35 12 37 6 37 e sin cos t x At B t t     A1 FT Must have “ x ”. FT on CF.   35 12 37 37 6 6 ' e 6e cos sin t t x A At B t t      M1* Differentiates using product rule. Must have their PI differentiated. 12 37 B  35 37 6 0 1 A B A      DM1 A1 Forms simultaneous equations using initial conditions.   35 12 12 37 37 37 6e sin cos t x t t t     A1 Must have “ x ”. 11

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7 (a) Use the substitution u = x 2 - 1 to find d x . [3] x 2 - 1 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ y O 1 2 3 N - 1 N x The diagram shows the curve with equation y = cosh -1 x together with a set of ( N - 1) rectangles of unit width. (b) By considering the sum of the areas of these rectangles, show that N ln r + r 2 - 1 2 N ln N + N 2 - 1 - N 2 - 1 . [5] / ` j ` j r = 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ N (c) Use a similar method to find, in terms of N, an upper bound for ln r + r 2 - 1 . 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Mark scheme: 7(a) 1 2 2 1 d d 1 x x u u x        2 1 x C  A1 Allow with “ C  ” missing. Answer must be in terms of x. 3 7(b)   1 2 cosh ln 1 r r r     B1 1 1 1 cosh 2 cosh 3 cosh N        M1 Forms sum of the areas of the rectangles. 1 1 cosh d N x x   M1 Compares with integral with correct limits. 1 1 2 1 1 1 cosh d cosh d 1 N N N x x x x x x x            A1 Evaluates integral.     2 2 2 2 ln 1 ln 1 1 N r r r N N N N          A1 AG. 5 Question Answer Marks Guidance 7(c) 1 1 1 (cosh 1) cosh 2 cosh ( 1) N         1 1 cosh d N x x   (or 1 2 cosh d N x x   ) M1 A1 Compares with integral with correct limits.     2 2 2 2 ln 1 ( 1)ln 1 1 N r r r N N N N           or       2 2 2 2 ln 1 ( 1)ln 1 1 2ln 2 3 3 N r r r N N N N             A1 Adds   2 ln 1 N N   to both sides. Alternative method for question 7(c) 1 1 1 1 (cosh 1) cosh 2 cosh ( 1) cosh N N             1 1 1 2 1 2 1 cosh d ln 1 1 N N x x x x x x                 (or 1 2 1 cosh d N x x    ) M1 A1 Compares with integral with correct limits.      2 2 2 2 ln 1 1 ln 1 2 2 N r r r N N N N N N           or      2 2 1 ln 1 2 2 2ln 2 3 3 N N N N N N          A1 Second alternative method for question 7(c) 1 1 1 cosh (1 1) cosh (2 1) cosh ( 1 1) N                1 2 1 1 2 cosh 1 d 1 ln 1 2 2 N N x x x x x x x x                 M1 A1 Compares with integral with correct limits.      2 2 1 ln 1 2 2 2ln 2 3 3 N N N N N N          A1 3

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8 (a) Starting from the definitions of sech and tanh in terms of exponentials, prove that 1 - sech2 t = tanh 2 t . 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The curve C has parametric equations + ln sech t, y = 1 + tanh 4 t, for t 2 0 . x = 12 tanh 2 t d y (b) Show that = - 4 sech2 t . [5] d x ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 y 9(c) Find the coordinates of the point on C with 2 = - , giving your answer in the form ( a + ln b, c) dx 2 where a, b and c are rational numbers. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 8(a) 2 sech e e t t t    e e tanh e e t t t t t                2 2 2 2 2 2 2 2 e e 4 e e 2 e 2 e 4 1 e e e e e e e e t t t t t t t t t t t t t t                            M1 A1 Expands, gets to   2 2 2 e 2 e 4 e e t t t t       for M1, AG. 3 8(b) 3 2 d 4tanh sech d y t t t  B1     2 2 3 d tanh sech tanh tanh sech 1 tanh d x t t t t t t t      M1 A1 M1 sensible attempt at derivative of x. 3 2 2 3 d d sech d d d 4tanh 4s t ech d tanh y y t t t t x t x      M1 A1 Applies chain rule, must substitute their d d y t and d d x t for M1, AG. 5 8(c) 2 2 2 2 3 2 h d a d d d h d d 8sec t n sech 8 d tanh ta h d n t t t x t y y t t x x t             M1 A1 Finds 2 2 d . d y x For M1 2 . d s d e d d ch tanh y c t t t x       AEF. Accept 2 8cosech .t    2 16 25 2 2 2 9 9 2 2 2 sech 8 8 1 tanh tanh tanh tanh t t t t t         M1 A1 Sets equal to 9 2 , uses identity from (a) or equivalent. Accept 16 9 2 sinh t  or 5 2 2 9 cosh . t     8 3 881 25 5 625 , ln , x y   A1 A1 A1 for each correct coordinate. ln3 t  6

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